Question

Figure shows a typical circuit for low-pass filter. An AC input $V_i$ = 10 mV is applied at the left end and the output $V_0$ is received at the right end. Find the output voltages for V = 10 k Hz, 100 kHz, 1.0 MHz and 10.0 MHz. Note that as the frequency is increased the output decreases and hence the name low-pass filter.

Answer 1

Here $V_1=10\times {10}^{-3}$ V,

R = $1\times {10}^3\Omega$

C = $10\times {10}^{-9}$ F

(a) $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

= $\frac{1}{2\pi \times 10\times {10}^3\times 10\times {10}^{-9}}$

= $\frac{1}{2\pi \times {10}^{-4}}=\frac{{10}^4}{2\pi }=\frac{5000}{\pi }$

Z = $\sqrt{R^2+X_c^2}$

= $\sqrt{(1+{10}^3)+(\frac{5000}{\pi }{)}^2}$

= $\sqrt{{10}^6+(\frac{5000}{\pi }{)}^2}$

$I_0=\frac{E_0}{Z}=\frac{V_1}{Z}$

= $\frac{10\times {10}^{-3}}{\sqrt{{10}^6+(\frac{5000}{\pi }{)}^2}}$

$V_0=I_0{X}_C$

= $\frac{{10}^{-2}}{\sqrt{{10}^5+\left(\frac{5000}{h^2}\right)}}\times \frac{5000}{\pi }$

= 0.00846 V = 8.46 mV = 8.5 mV

$X_c=\frac{1}{\omega C}=\frac{1}{2\pi fc}$

= $\frac{1}{2\pi \times {10}^5\times 10\times {10}^{-9}}$

= $\frac{1}{2\pi {10}^{-3}}$

= $\frac{{10}^3}{2\pi }=\frac{1000}{2\pi }=\frac{500}{\pi }$

Z = $\sqrt{R^2+X_2^c}$

= $\sqrt{({10}^3{)}^2+(\frac{500}{\pi }{)}^2}$

$I_0=\frac{E_0}{Z}=\frac{V_1}{Z}$

= $\frac{10\times {10}^3}{\sqrt{{10}^5+(\frac{500}{\pi }{)}^2}}$

$V_0=I_0{X}_c$

= $\frac{{10}^2}{\sqrt{{10}^5+(\frac{50}{\pi }{)}^2}}\times \frac{500}{\pi }$

= 1.6124 V = 1.6 mV

(c) f = 1 MHz = ${10}^6$ Hz

$X_c=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

= $\frac{1}{2\pi \times {10}^6\times {10}^{-9}\times 10}$

= $\frac{1}{2\pi \times {10}^{-2}=\frac{100}{2\pi }=\frac{500}{\pi }}$

Z= $\sqrt{R^2+X^{2}C}$

= $\sqrt{({10}^3{)}^2+(\frac{50}{\pi }{)}^2}$

$I_0=\frac{E_0}{Z}=\frac{V_1}{Z}$

=$\frac{10\times {10}^3}{{10}^5+(\frac{500}{\pi }{)}^2}$

$V_0=I_0{X}_c$

=$\frac{{10}^2}{\sqrt{{10}^5+(\frac{50}{\pi }{)}^2}}\times \frac{500}{\pi }$

= 1.6124 V = 1.6 mV

(c) f= 1 MHz = ${10}^6$ Hz

$X_c=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

=$\frac{1}{2\pi \times {10}^6\times {10}^{-9}\times 10}$

=$\frac{1}{2\pi \times {10}^{-2}}=\frac{100}{2\pi }=\frac{500}{\pi }$

Z = $\sqrt{R^2+X^{2}C}$

= $\sqrt{({10}^3{)}^2+(\frac{50}{\pi }{)}^2}$

$I_0=\frac{E_0}{Z}=\frac{V_1}{Z}$

= $\frac{10\times {10}^{-3}}{\sqrt{{10}^6+(\frac{50}{\pi }{)}^2}}$

$V_0{I}_0{X}_c$

=$\frac{{10}^{-2}}{\sqrt{{10}^5+(\frac{50}{\pi }{)}^2}}\times \frac{50}{\pi }$

= 0.16 mV

(d) f = 10 MHz

= $10\times {10}^{6}Hz={10}^7$ Hz

$X_c=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

= $\frac{1}{2\pi \times {10}^7\times 10\times {10}^{-9}}=\frac{5}{\pi }$

Z =$\sqrt{R^2+X_c^2}$

= $\sqrt{({10}^3{)}^2+(\frac{5}{\pi }{)}^2}$

$i_0=\frac{E_0}{2}=\frac{V_1}{Z}$

= $\frac{10\times {10}^{-3}}{\sqrt{{10}^6+(\frac{5}{\pi }{)}^2}}$

$V_0=I_0{X}_c$

= $\frac{{10}^{-2}}{\sqrt{{10}^6+\frac{5}{\pi }{)}^2}}\times \frac{5}{\pi }=16\mu$ V.